Reflection Coefficient - Handout Assignment | POLI 10, Assignments of Political Science

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Part C The plots in this section are nice to look at but you don’t actually need them! Note that at the desired wavelength, along with the given thickness and refractive index of the SiO 2 layers, allows for exactly an odd multiple of wavelengths to fit inside those layers. The quarter-wave stack (i.e. a stack of different materials with thicknesses each measuring an odd multiple of a quarter of the wavelength of interest) is the simplest way to manufacture a high-reflective coating. So to get peak reflection in our problem, we should choose d 3 to be an odd multiple of one-fourth of 500 nm. I’ll choose 125 nm; however, note that the plots are periodic every 250 nm (as they should be!). You should also notice that, beginning at zero and going with the periodicity of the plot, you are able to achieve unity transmission. Half-wavelength layers have historically been called latent layers since light traverses these layers without suffering any reflection as if the material weren’t there! Great, but we still have the glass, shouldn’t there still be some reflection? Well each of the SiO 2 layers represent quarter- wavelength layers; by virtue of having a latent layer in the middle, we effectively bring the two panes of glass together and essentially make another latent, half-wavelength layer. Part D At normal incidence, we’ve designed the stack to give a maximum in reflection at 500 nm. By plotting the coefficients as a function of wavelength, we see that this is indeed the case. We also see that other extrema exist; these are points at which we also meet the quarter- or half-wavelength conditions previously discussed. Note that the widths of the peaks and valleys get wider for longer wavelengths indicating that the interference effects become less effective. This makes sense since, for longer waves, a full wavelength may not even be able to fit inside some of the layers. t r

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By plotting the 500 nm case as a function of incident angle, we see that the results are much richer than the rather dull case of reflection/transmission from a single interface. Whereas the s-polarized intensity coefficients in the single interface case are monotonic functions of the incident angle, we see this is no longer true with multiple interfaces. Indeed, there is a point (or rather band) where the s-wave suffers no reflection. This effect, however, is not really a Brewster angle but rather due to changes in the optical path lengths that resemble the half-wavelength condition. It is likely, however, that the wider unity transmission band and/or the second reflection minimum in the p-polarized intensity plot is due, at least partly, to a Brewster effect. Part E Unlike the setup from Part C, the slightly more complicated setup here makes it a little trickier to foresee what will happen before making the plots. From the dispersion given for gold, we choose n=0.1033 ± i1.942. Note that, for either real or complex values of the refractive index and permittivity, ! /! 0 =n^2 (non-permeable media) and that the choice of the sign of the imaginary part will be determined by your choice of coordinates (i.e. whether your choice of forward wave propagation is in the + or – direction). In this setup, although we clearly have reflection/transmission extrema, there is no point where we have neither unity transmission nor zero reflection. Additionally, the sum of reflection and transmission intensity coefficients is always less than unity. These three effects are due to the loss introduced by the metal layer. Note that the plots are still periodic every 250 nm (as they should be!). I’ll choose 212 nm for the third layer. t r r

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